FFT function
On 10/04/2009, at 11:44 PM, Achilleas Achilleos wrote:
Hi, A very simple question. I know that the Fourier transform of a Laplace distribution, with zero mean and variance 2b^2, is equal to 1/(1+(tb)^2). Therefore, the Fourier transform is a positive function (actually is always between 0 and 1). BUT, if I use the fft(alpha) function of R, where alpha is a vector containing 128 values of the probability density function from -2 to 2, I get negative values as well. Why?
Basically because there is a (considerable!) difference between the
continuous
Fourier transform and the discrete Fourier transform, the latter
being what
is calculated by fft(). (NOTE: ***NOT*** ``FFT'' --- R is case
sensitive.)
cheers,
Rolf Turner
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